EXAMPLE 1 Solve an equation with two real solutions

Solve x2 + 3x = 2.x2 + 3x = 2 Write original equation.

x2 + 3x – 2 = 0 Write in standard form.

x =– b + b2 – 4ac2a

Quadratic formula

x =– 3 + 32 – 4(1)(–2) 2(1)

a = 1, b = 3, c = –2

Simplify.x = – 3 + 172

The solutions are x = – 3 + 172

0.56 andx = – 3 – 17

2– 3.56.

ANSWER

EXAMPLE 1 Solve an equation with two real solutions

CHECK

Graph y = x2 + 3x – 2 and note that the x-intercepts are about 0.56 and about – 3.56.

EXAMPLE 2 Solve an equation with one real solutions

Solve 25x2 – 18x = 12x – 9.25x2 – 18x = 12x – 9. Write original equation.

Write in standard form.

x =30 + (–30)2– 4(25)(9)2(25)

a = 25, b = –30, c = 9

Simplify.

25x2 – 30x + 9 = 0.

x =30 + 0

50

x = 35 Simplify.

35The solution is

ANSWER

EXAMPLE 2 Solve an equation with one real solutions

CHECK

Graph y = –5x2 – 30x + 9 and note that the only x-intercept is 0.6 = .3

5

EXAMPLE 3 Solve an equation with imaginary solutions

Solve –x2 + 4x = 5.–x2 + 4x = 5 Write original equation.

Write in standard form.

x =– 4+ 42– 4(– 1)(– 5)2(– 1)

a = –1, b = 4, c = –5

Simplify.

–x2 + 4x – 5 = 0.

x =– 4+ – 4

– 2– 4+ 2i

x = – 2

Simplify.

Rewrite using the imaginary unit i.

x = 2 + i

The solution is 2 + i and 2 – i.

ANSWER

EXAMPLE 3 Solve an equation with imaginary solutions

CHECK

Graph y = 2x2 + 4x – 5. There are no x-intercepts. So, the original equation has no real solutions. The algebraic check for the imaginary solution 2 + i is shown.

–(2 + i)2 + 4(2 + i) = 5?

–3 – 4i + 8 + 4i = 5?

5 = 5

EXAMPLE 4 Use the discriminant

Find the discriminant of the quadratic equation and give the number and type of solutions of the equation.

a. x2 – 8x + 17 = 0 b. x2 – 8x + 16 = 0 c. x2 – 8x + 15 = 0

SOLUTION

Equation Discriminant Solution(s)

ax2 + bx + c = 0 b2 – 4ac x =– b+ b2– 4ac2ac

a. x2 – 8x + 17 = 0 (–8)2 – 4(1)(17) = – 4 Two imaginary: 4 + i

b. x2 – 8x + 16 = 0 (–8)2 – 4(1)(16) = 0 One real: 4

b. x2 – 8x + 15 = 0 (–8)2 – 4(1)(15) = 0 Two real: 3,5

EXAMPLE 1 Graph a quadratic inequality

Graph y > x2 + 3x – 4.

SOLUTION

STEP 1

Graph y = x2 + 3x – 4. Because the inequality symbol is >, make the parabola dashed.

Test a point inside the parabola, such as (0, 0).

STEP 2

y > x2 + 3x – 4

0 > 02 + 3(0) – 4?

0 > – 4

EXAMPLE 1 Graph a quadratic inequality

So, (0, 0) is a solution of the inequality.

STEP 3

Shade the region inside the parabola.

EXAMPLE 2 Use a quadratic inequality in real life

A manila rope used for rappelling down a cliff can safely support a weight W (in pounds) provided

Rappelling

W ≤ 1480d2

where d is the rope’s diameter (in inches). Graph the inequality.

SOLUTION

Graph W = 1480d2 for nonnegative values of d. Because the inequality symbol is ≤, make the parabola solid. Test a point inside the parabola, such as (1, 2000).

EXAMPLE 2 Use a quadratic inequality in real life

W ≤ 1480d2

2000 ≤ 1480(1)2

2000 ≤ 1480

Because (1, 2000) is not a solution, shade the region below the parabola.

EXAMPLE 3 Graph a system of quadratic inequalities

Graph the system of quadratic inequalities.

y < – x2 + 4 Inequality 1

y > x2 – 2x – 3 Inequality 2

SOLUTION

STEP 1

Graph y ≤ – x2 + 4. The graph is the red region inside and including the parabola y = – x2 + 4.

EXAMPLE 3 Graph a system of quadratic inequalities

STEP 2

Graph y > x2– 2x – 3. The graph is the blue region inside (but not including) the parabola y = x2 –2x – 3.

Identify the purple region where the two graphs overlap. This region is the graph of the system.

STEP 3

EXAMPLE 4 Solve a quadratic inequality using a table

Solve x2 + x ≤ 6 using a table.

SOLUTION

Rewrite the inequality as x2 + x – 6 ≤ 0. Then make a table of values.

Notice that x2 + x –6 ≤ 0 when the values of x are between –3 and 2, inclusive.

The solution of the inequality is –3 ≤ x ≤ 2.

ANSWER

EXAMPLE 5 Solve a quadratic inequality by graphing

Solve 2x2 + x – 4 ≥ 0 by graphing.

SOLUTION

The solution consists of the x-values for which the graph of y = 2x2 + x – 4 lies on or above the x-axis. Find the graph’s x-intercepts by letting y = 0 and using the quadratic formula to solve for x.

0 = 2x2 + x – 4

x =– 1+ 12– 4(2)(– 4)2(2)

x =– 1+ 33

4x 1.19 or x –1.69

EXAMPLE 5 Solve a quadratic inequality by graphing

Sketch a parabola that opens up and has 1.19 and –1.69 as x-intercepts. The graph lies on or above the x-axis to the left of (and including) x = – 1.69 and to the right of (and including) x = 1.19.

The solution of the inequality is approximately x ≤ – 1.69 or x ≥ 1.19.

ANSWER

EXAMPLE 1 Write a quadratic function in vertex form

Write a quadratic function for the parabola shown.

SOLUTION

Use vertex form because the vertex is given.

y = a(x – h)2 + k Vertex form

y = a(x – 1)2 – 2 Substitute 1 for h and –2 for k.

Use the other given point, (3, 2), to find a.2 = a(3 – 1)2 – 2 Substitute 3 for x and 2 for y.

2 = 4a – 2 Simplify coefficient of a.

1 = a Solve for a.

EXAMPLE 1 Write a quadratic function in vertex form

A quadratic function for the parabola is y = (x – 1)2 – 2.

ANSWER

EXAMPLE 2 Write a quadratic function in intercept form

Write a quadratic function for the parabola shown.

SOLUTION

Use intercept form because the x-intercepts are given.

y = a(x – p)(x – q) Intercept form

y = a(x + 1)(x – 4) Substitute –1 for p and 4 for q.

EXAMPLE 2 Write a quadratic function in intercept form

Use the other given point, (3, 2), to find a.

2 = a(3 + 1)(3 – 4) Substitute 3 for x and 2 for y.

2 = – 4a Simplify coefficient of a.

Solve for a.12

– = a

A quadratic function for the parabola is

12

– (x + 1)(x – 4) .y =

ANSWER

EXAMPLE 3 Write a quadratic function in standard form

Write a quadratic function in standard form for the parabola that passes through the points (–1, –3), (0, – 4), and (2, 6).

SOLUTION

STEP 1

Substitute the coordinates of each point into y = ax2 + bx + c to obtain the system of three linear equations shown below.

EXAMPLE 3 Write a quadratic function in standard form

–3 = a(–1)2 + b(–1) + c Substitute –1 for x and 23 for y.

–3 = a – b + c Equation 1

–3 = a(0)2 + b(0) + c Substitute 0 for x and – 4 for y.

– 4 = c Equation 2

6 = a(2)2 + b(2) + c Substitute 2 for x and 6 for y.

6 = 4a + 2b + c Equation 3

Rewrite the system of three equations in Step 1 as a system of two equations by substituting – 4 for c in Equations 1 and 3.

STEP 2

EXAMPLE 3 Write a quadratic function in standard form

a – b + c = – 3 Equation 1

a – b – 4 = – 3 Substitute – 4 for c.

a – b = 1 Revised Equation 1

4a + 2b + c = 6 Equation 3

4a + 2b + 4 = 6 Substitute – 4 for c.

4a + 2b = 10 Revised Equation 3

STEP 3Solve the system consisting of revised Equations 1 and 3. Use the elimination method.

EXAMPLE 3 Write a quadratic function in standard form

a – b = 1 2a – 2b = 2

4a + 2b = 10 4a + 2b = 10

6a = 12

a = 2

So 2 – b = 1, which means b = 1.

The solution is a = 2, b = 1, and c = – 4.

A quadratic function for the parabola is y = 2x2 + x – 4.

ANSWER